Given,

F(1) = 0, F(3) = 4

F'(x) < 0 for all x $$\in$$(1, 3)

and f(x) = xF(x)

Now, f'(x) = F(x) + xF'(x)

$$\Rightarrow$$ f'(1) = F(1) + 1 . F'(1)

$$\Rightarrow$$ f'(1) = 0 + F'(1) ....... (1)

As F'(x) < 0 for all x $$\in$$ (1, 3)

$$\therefore$$ F'(1) < 0

From (1), we get

f'(1) = F'(1) < 0

From Lagrange theorem

$$F'(2) = {{F(3) - F(1)} \over {3 - 1}}$$

$$ = {{ - 4 - 0} \over 2}$$

= $$-$$2

As f(x) = xF(x)

$$\therefore$$ f(3) = 3 . F(3) = 3 . ($$-$$4) = $$-$$12

and f(1) = 1 . F(1) = 0

$$\therefore$$ $$f'(2) = {{f(3) - f(1)} \over {3 - 1}}$$

$$ = {{ - 12} \over 2} = - 6$$

As, f'(x) = F(x) + x . F'(x)

$$\therefore$$ f'(2) = F(2) + 2 . F'(2)

$$ \Rightarrow - 6 = {{f(2)} \over 2} + 2.\,( - 2)$$

$$ \Rightarrow {{f(2)} \over 2} = - 2$$

$$\Rightarrow$$ f(2) = $$-$$4

$$\therefore$$ f(2) < 0

$$f'(x) = {{f(3) - f(1)} \over {3 - 1}}$$ when x $$\in$$ (1, 3)

f(3) = $$-$$12

and f(1) = 0

$$\therefore$$ $$f'(x) = {{ - 12} \over 2} = - 6$$

$$\therefore$$ f'(x) $$\ne$$ 0 for any x $$\in$$(1, 3)

Given, $$\int_1^3 {{x^2}F'(x)dx = - 12} $$

$$ \Rightarrow [{x^2}F(x)]_1^3 - \int_1^3 {2x\,.\,F(x)dx = - 12} $$

$$ \Rightarrow 9F(3) - F(1) - 2\int_1^3 {f(x)dx = - 12} $$

$$ \Rightarrow - 36 - 0 - 2\int_1^3 {f(x)dx = - 12} $$

$$\therefore$$ $$\int_1^3 {f(x)dx = - 12} $$

and $$\int_1^3 {{x^3}F''(x)dx = 40} $$

$$ \Rightarrow [{x^3}F'(x)]_1^3 - \int_1^3 {3{x^2}F'(x)dx = 40} $$

$$ \Rightarrow [{x^2}(xF'(x)]_1^3 - 3 \times ( - 12) = 40$$

$$ \Rightarrow \{ {x^2}\,.\,[f'(x) - F(x)]\} _1^3 = 4$$

$$ \Rightarrow 9[f'(3) - F(3)] - [f'(1) - F(1)] = 4$$

$$ \Rightarrow 9[f'(3) + 4] - [f'(1) - 0] = 4$$

$$ \Rightarrow 9f'(3) - f'(1) = - 32$$

Let $${I_1} = \int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = \int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } } } $$

$$\therefore$$ $${I_1} = {I_2} + {I_3} + {I_4} + {I_5}$$ ...... (i)

Now, $${I_3} = \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

Put $$t = \pi + t \Rightarrow dt = dt$$

$$\therefore$$ $${I_3} = \int_0^\pi {{e^{\pi + t}}.\,({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = {e^t}\,.\,{I_2}$$ ..... (ii)

Now, $${I_4} = \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

Put $$t = 2\pi + t \Rightarrow dt = dt$$

$$\therefore$$ $${I_4} = \int_0^\pi {{e^{t + 2\pi }}({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = {e^{2\pi }}.{I_2}$$ ...... (iii)

and $${I_5} = \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

Put $$t = 3\pi + t$$

$$\therefore$$ $${I_5} = \int_0^\pi {{e^{3\pi + t}}({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = {e^{3\pi }}.\,{I_2}$$ ...... (iv)

From Eqs. (i), (ii), (iii) and (iv), we get

$${I_1} = {I_2} + {e^\pi }.\,{I_2} + {e^{2\pi }}.\,{I_2} + {e^{3\pi }}.\,{I_2}$$

$$ = (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }}){I_2}$$

$$\therefore$$ $$L = {{\int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } \over {\int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} }}$$

$$ = (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }})$$

$$ = {{1.\,({e^{4\pi }} - 1)} \over {{e^\pi } - 1}}$$ for $$a \in R$$

$$f(x) = 7{\tan ^8}x + 7{\tan ^6}x - 3{\tan ^4}x - 3{\tan ^2}x\,\forall x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)$$

$$ = 7{\tan ^6}x\,.\,{\sec ^2}x - 3{\tan ^2}x\,.\,{\sec ^2}x$$

$$ = (7{\tan ^6}x - 3{\tan ^2}x)\,.\,{\sec ^2}x$$

$$ \Rightarrow \int\limits_0^{\pi /4} {f(x)dx = \int\limits_0^{\pi /4} {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}dx = \int\limits_0^1 {(7 + 603{t^2})dt = [{t^7} - {t^3}]_0^1 = 0} } } $$

Also, $$I = \int\limits_0^{\pi /4} {xf(x)dx} $$

$$ = \left| {x\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}xdx} } \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {1\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}x\,dx} } $$

$$ = \left| {x\,.\,({{\tan }^7}x - {{\tan }^3}x} \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {({{\tan }^7}x - {{\tan }^3}x)dx} $$

$$ = 0 - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^4}x - 1)dx} $$

$$ = - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^2}x - 1)({{\sec }^2}x)dx} $$

$$ = - \int\limits_0^1 {({t^5} - {t^3})dt} $$

$$ = - \left[ {{{{t^6}} \over 6} - {{{t^4}} \over 4}} \right]_0^1 = \left[ {{1 \over 4} - {1 \over 6}} \right] = {1 \over {12}}$$